Page:Annales de mathématiques pures et appliquées, 1821-1822, Tome 12.djvu/288

280

QUADRILATÈRE

${\begin{aligned}&\operatorname {Sin} .{\frac {1}{4}}T=\\&{\sqrt {\frac {\operatorname {Sin} .{\frac {1}{4}}(a+b+c)\operatorname {Sin} .{\frac {1}{4}}(b+c-a)\operatorname {Sin} .{\frac {1}{4}}(c+a-b)\operatorname {Sin} .{\frac {1}{4}}(a+b-c)}{\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}},\\\\&\operatorname {Cos} .{\frac {1}{4}}T=\\&{\sqrt {\frac {\operatorname {Cos} .{\frac {1}{4}}(a+b+c)\operatorname {Cos} .{\frac {1}{4}}(b+c-a)\operatorname {Cos} .{\frac {1}{4}}(c+a-b)\operatorname {Cos} .{\frac {1}{4}}(a+b-c)}{\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}},\\\\&\operatorname {Sin} .{\frac {1}{2}}T=\\&{\sqrt {\frac {\operatorname {Sin} .{\frac {1}{2}}(a+b+c)\operatorname {Sin} .{\frac {1}{2}}(b+c-a)\operatorname {Sin} .{\frac {1}{2}}(c+a-b)\operatorname {Sin} .{\frac {1}{2}}(a+b-c)}{2\operatorname {Cos} .{\frac {1}{2}}a\operatorname {Cos} .{\frac {1}{2}}b\operatorname {Cos} .{\frac {1}{2}}c}}},\\\\&\operatorname {Tang} .{\frac {1}{4}}T=\\&{\sqrt {\operatorname {Tang} .{\tfrac {1}{4}}(a+b+c)\operatorname {Tang} .{\tfrac {1}{4}}(b+c-a)\operatorname {Tang} .{\tfrac {1}{4}}(c+a-b)\operatorname {Tang} .{\tfrac {1}{4}}(a+b-c)}}\,;\end{aligned}}$

formules connues, dont la dernière est due à M. Lhuilier, de Genève.

Si nous désignons par $R$ l’arc de grand cercle qui joint le pôle du cercle auquel notre quadrilatère est inscrit à l’un quelconque des points de sa circonférence, cet arc aura pour sinus le rayon même de ce cercle, de sorte que, pour obtenir $\operatorname {Sin} .R,$ il ne s’agira que de changer, dans la formule (5) précédemment obtenue,

$a,b,c,d,R$

respectivement en

$2\operatorname {Sin} .{\tfrac {1}{2}}a,\ 2\operatorname {Sin} .{\tfrac {1}{2}}b,\ 2\operatorname {Sin} .{\tfrac {1}{2}}c,\ 2\operatorname {Sin} .{\tfrac {1}{2}}d,\ \operatorname {Sin} .R$

Posant donc, pour abréger,

${\begin{aligned}\operatorname {Sin} .{\tfrac {1}{2}}b+\operatorname {Sin} .{\tfrac {1}{2}}c+\operatorname {Sin} .{\tfrac {1}{2}}d-\operatorname {Sin} .{\tfrac {1}{2}}a&=\operatorname {Sin} .A,\\\operatorname {Sin} .{\tfrac {1}{2}}c+\operatorname {Sin} .{\tfrac {1}{2}}d+\operatorname {Sin} .{\tfrac {1}{2}}a-\operatorname {Sin} .{\tfrac {1}{2}}b&=\operatorname {Sin} .B,\\\operatorname {Sin} .{\tfrac {1}{2}}d+\operatorname {Sin} .{\tfrac {1}{2}}a+\operatorname {Sin} .{\tfrac {1}{2}}b-\operatorname {Sin} .{\tfrac {1}{2}}c&=\operatorname {Sin} .C,\\\operatorname {Sin} .{\tfrac {1}{2}}a+\operatorname {Sin} .{\tfrac {1}{2}}b+\operatorname {Sin} .{\tfrac {1}{2}}c-\operatorname {Sin} .{\tfrac {1}{2}}d&=\operatorname {Sin} .D\,;\\\\\end{aligned}}$

$\left(\operatorname {Sin} .{\tfrac {1}{2}}a\operatorname {Sin} .{\tfrac {1}{2}}b+\operatorname {Sin} .{\tfrac {1}{2}}c\operatorname {Sin} .{\tfrac {1}{2}}d\right)\left(\operatorname {Sin} .{\tfrac {1}{2}}a\operatorname {Sin} .{\tfrac {1}{2}}c+\operatorname {Sin} .{\tfrac {1}{2}}b\operatorname {Sin} .{\tfrac {1}{2}}d\right)\times$

$\left(\operatorname {Sin} .{\frac {1}{2}}b\operatorname {Sin} .{\frac {1}{2}}c+\operatorname {Sin} .{\frac {1}{2}}a\operatorname {Sin} .{\frac {1}{2}}d\right)=\operatorname {Sin} .K,$

il viendra

$\operatorname {Sin} .R=2{\sqrt {\frac {\operatorname {Sin} .K}{\operatorname {Sin} .A\operatorname {Sin} .B\operatorname {Sin} .C\operatorname {Sin} .D}}}.\qquad$(V)

Il est presque superflu d’observer que les résultats que nous venons d’obtenir, en dernier lieu, s’appliquent littéralement à l’angle tétraèdre inscrit au cône droit.

Dijon, le 20 décembre 1821.